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$18 \mathrm{~g}$ of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ is added to $178.2 \mathrm{~g}$ of water. The vapour pressure of water for this aqueous solution at $100^{\circ} \mathrm{C}$ is
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The correct answer is:
$752.40$ Torr
$752.40$ Torr
$$
\begin{aligned}
& \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{\mathrm{s}}}=\frac{\mathrm{n}}{\mathrm{N}} \\
& \frac{760-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{\mathrm{s}}}=\frac{\frac{18}{180}}{\frac{178.2}{18}}=\frac{\frac{1}{10}}{9.9}=\frac{0.1}{9.9} \\
& 760-\mathrm{P}_{\mathrm{s}}=\frac{1}{99} \mathrm{P}_{\mathrm{s}} \\
& 760 \times 99-\mathrm{P}_{\mathrm{s}} \times 99=\mathrm{P}_{\mathrm{s}} \\
& 760 \times 99=100 \mathrm{P}_{\mathrm{s}} \\
& \mathrm{P}_{\mathrm{s}}=\frac{760 \times 99}{100}=752.4
\end{aligned}
$$
\begin{aligned}
& \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{\mathrm{s}}}=\frac{\mathrm{n}}{\mathrm{N}} \\
& \frac{760-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{\mathrm{s}}}=\frac{\frac{18}{180}}{\frac{178.2}{18}}=\frac{\frac{1}{10}}{9.9}=\frac{0.1}{9.9} \\
& 760-\mathrm{P}_{\mathrm{s}}=\frac{1}{99} \mathrm{P}_{\mathrm{s}} \\
& 760 \times 99-\mathrm{P}_{\mathrm{s}} \times 99=\mathrm{P}_{\mathrm{s}} \\
& 760 \times 99=100 \mathrm{P}_{\mathrm{s}} \\
& \mathrm{P}_{\mathrm{s}}=\frac{760 \times 99}{100}=752.4
\end{aligned}
$$
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